Question

Wooden artifact and freshly cut tree are ${\text{7}}{\text{.6}}$ and $15.2{\text{ mi}}{{\text{n}}^{ - 1}}{\text{ }}{{\text{g}}^{ - 1}}$ of carbon $\left( {{t_{{\text{1/2}}}} = 5760{\text{ years}}} \right)$ respectively. The age of the artifact is:
A) $5760{\text{ years}}$
B) $5760 \times \dfrac{{15.2}}{{7.6}}{\text{ years}}$
C) $5760 \times \dfrac{{7.6}}{{15.2}}{\text{ years}}$
D) $5760 \times \left( {15.2 - 7.6} \right){\text{ years}}$

Answer 1

To solve this we must know the equation to calculate the age of a radioactive nuclide. In the equation, substitute the values given and solve for the age of the artifact. Remember the units of the half-life of the nuclide and the time taken for the nuclide to decay are the same.

Complete step by step solution:
The equation to calculate the age of a radioactive nuclide is given by the equation as follows:
${N_t} = {N_0}{\left( {{\text{1/2}}} \right)^{t/{t_{_{1/2}}}}}$
Where ${N_t}$ is the amount of nuclide left at time $t$,
${N_0}$ is the initial amount of nuclide,
$t$ is the time taken for ${N_0}$ to go to ${N_t}$,
${t_{{\text{1/2}}}}$ is the half-life of the nuclide.
We are given that a wooden artifact and freshly cut tree are ${\text{7}}{\text{.6}}$ and $15.2{\text{ mi}}{{\text{n}}^{ - 1}}{\text{ }}{{\text{g}}^{ - 1}}$ of carbon $\left( {{t_{{\text{1/2}}}} = 5760{\text{ years}}} \right)$. Thus,
${\text{7}}{\text{.6 mi}}{{\text{n}}^{ - 1}}{\text{ }}{{\text{g}}^{ - 1}} = 15.2{\text{ mi}}{{\text{n}}^{ - 1}}{\text{ }}{{\text{g}}^{ - 1}}{\left( {{\text{1/2}}} \right)^{{\text{t/}}5760{\text{ years}}}}$
$0.5 = {\left( {{\text{1/2}}} \right)^{{\text{t/}}5760{\text{ years}}}}$
$1 = {1^{{\text{t/}}5760{\text{ years}}}}$
$t = 5760{\text{ years}}$
Thus, the age of the artifact is $5760{\text{ years}}$.
Thus, the correct option is (A) $5760{\text{ years}}$.

Note: Remember the units of the half-life of the nuclide and the time taken for ${N_0}$ to go to ${N_t}$ are the same. Time can be measured in seconds, minutes, days, weeks, months or years. The units of the half-life of the nuclide and the time taken for ${N_0}$ to go to ${N_t}$ i.e. ${t_{{\text{1/2}}}}$ and $t$ remain same.